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函数y=f(x)=ax^2+2ax+4(0
人气:125 ℃ 时间:2020-02-03 19:19:49
解答
应该是0

f(x2)-f(x1)=a(x2^2-x1^2)+2a(x2-x1)
=a(x2+x1)(x2-x1)+2a(x2-x1)
=a(1-a)(x2-x1)+2a(x2-x1)
=-a(a-3)(x2-x1)
∵0x1
∴f(x2)-f(x1)>0
f(x2)>f(x1)
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